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Math Help - find the distance of a point from a line?

  1. #1
    Newbie Miasmagasma's Avatar
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    find the distance of a point from a line?

    i'm truly sorry if this is the wrong forum, as i'm teaching myself this stuff, i'm unsure of what it's sub forum should be. since my last question was moved here, i put this one here too.

    anyway.....

    use the following procedure to find the leats ( perpendicular ) distance of the point ( 1,2 ) from the line y = 3x + 5, without having to find the equation of a line perpendicular to y = 3x + 5.

    (a) let ( x,y ) be a general point on the line, Show that its distance, d, from ( 1,2 ) is given by d^2 = ( x - 1 )^2 + ( y - 2 )^2.

    (b) use the equation of the line to show that d^2 = ( x - 1 )^2 + ( 3x 3 )^2. I CAN SEE THE ANSWER HERE SO NO WORRIES.

    (c) by completing the square, show that the minimum distance required 3/5√10
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  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by Miasmagasma View Post
    (b) use the equation of the line to show that d^2 = ( x - 1 )^2 + ( 3x + 3 )^2. I CAN SEE THE ANSWER HERE SO NO WORRIES.

    (c) by completing the square, show that the minimum distance required 3/5√10
    d^2 = ( x - 1 )^2 + ( 3x +  3 )^2
    = 10x^2 + 16x + 10
    = (\sqrt{10}x)^2 + 2\left(\frac{8}{\sqrt{10}}\right)(\sqrt{10}x) + \left(\frac{8}{\sqrt{10}}\right)^2 + \left(10 - \frac{64}{10}\right)
    = \left(\sqrt{10}x + \frac{8}{\sqrt{10}}\right)^2 + \frac{36}{10}
    \boxed{d^2 \geq \frac{36}{10}} since \left(\sqrt{10}x + \frac{8}{\sqrt{10}}\right)^2 \geq 0

    d^2 \geq \frac{36}{10} \implies d \geq \frac{6}{\sqrt{10}}

    Now show that \frac{6}{\sqrt{10}} = \frac{3\sqrt{10}}{5} and you are done.
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  3. #3
    Newbie Miasmagasma's Avatar
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    Quote Originally Posted by Isomorphism View Post
    d^2 = ( x - 1 )^2 + ( 3x +  3 )^2
    = 10x^2 + 16x + 10
    = (\sqrt{10}x)^2 + 2\left(\frac{8}{\sqrt{10}}\right)(\sqrt{10}x) + \left(\frac{8}{\sqrt{10}}\right)^2 + \left(10 - \frac{64}{10}\right)
    = \left(\sqrt{10}x + \frac{8}{\sqrt{10}}\right)^2 + \frac{36}{10}
    \boxed{d^2 \geq \frac{36}{10}} since \left(\sqrt{10}x + \frac{8}{\sqrt{10}}\right)^2 \geq 0

    d^2 \geq \frac{36}{10} \implies d \geq \frac{6}{\sqrt{10}}

    Now show that \frac{6}{\sqrt{10}} = \frac{3\sqrt{10}}{5} and you are done.
    can you explain how you changed it to this...



    ?
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by Miasmagasma View Post
    can you explain how you changed it to this...



    ?
    That step is generally called completing the square. Read about it here.
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  5. #5
    Newbie Miasmagasma's Avatar
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    Quote Originally Posted by Isomorphism View Post
    That step is generally called completing the square. Read about it here.
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  6. #6
    Newbie Miasmagasma's Avatar
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    i haven't been taught that method of completing the square in my book. is it anything to do with it being OCR board?

    i have only learnt the ( x + 1/2b )^2 - 1/4b^2 + c

    can that method be used as well?
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  7. #7
    Newbie Miasmagasma's Avatar
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    could someone please walk me through this? i'm losing my marbles. i attempted the problem but i have no idea why he has square rooted all the 10s. if you are meant to divide through the whole equation why isn't it

    x^2 + 16/10x + 1?????????????
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  8. #8
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Miasmagasma View Post
    could someone please walk me through this? i'm losing my marbles. i attempted the problem but i have no idea why he has square rooted all the 10s. if you are meant to divide through the whole equation why isn't it

    x^2 + 16/10x + 1?????????????
    \bigg(x+\frac{8}{5}\bigg)^2 +1 -\frac{64}{100}

    \bigg(x+\frac{8}{5}\bigg)^2 +\frac{36}{100}

    Now remember you divided by 10 at the begining so you have to multiply 10 back in

    10\bigg(x+\frac{8}{5}\bigg)^2 +(10)\frac{36}{100}

    so now you have

    d^2 = \bigg({10\bigg(x+\frac{8}{5}\bigg)^2 +(10)\frac{36}{100}}\bigg)^2

    To solve for d take square root of both sides

    d = \sqrt{10\bigg(x+\frac{8}{5}\bigg)^2 +(10)\frac{36}{100}}
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